symmetry methods for differential equations,exercise 1.4

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	itle{	extbf{Symmetry Methods for Differential Equations:\Exercise 1.4}} 
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author{small{叶卢庆}\{small{杭州师范大学理学院,学号:1002011005}}\{small{Email:h5411167@gmail.com}}} % Institution

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egin{exercise}[1.4]
Determine the value of $alpha$ for which 
$$
(x',y')=(x+2va,ye^{alphava})
$$
is a symmetry of 
$$
frac{dy}{dx}=y^2e^{-x}+y+e^x
$$
for all $vainmathbf{R}$.  
end{exercise}
egin{proof}
  The symmetry condition for the differential equation is 
$$
frac{frac{pa g}{pa x}+frac{pa g}{pa y}w(x,y)}{frac{pa f}{pa
    x}+frac{pa f}{pa y}w(x,y)}=w(f(x,y),g(x,y)).
$$
Where
$w(x,y)=y^2e^{-x}+y+e^x$,$f(x,y)=x+2va,g(x,y)=ye^{alphava}$.So the
symmetry condition can be written as 
$$
y^2e^{-x+alphava}+e^{x+alphava}=y^2e^{2alphava-x-2va}+e^{x+2va}.
$$
So $alpha=2$.
end{proof}
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